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SSC CGL solved paper Tier-I exam held on 16 August 2015 (II-Shift): Quantitative Aptitude

Aug 3, 2018 20:19 IST
    SSC CGL solved paper
    SSC CGL solved paper

    In this article, we are providing 50 solved questions of Aptitude section asked in SSC CGL tier-1 exam held on 16thAugust, 2015 (Evening shift). Please find the topic wise questions’ distribution in the following table- 

    From the above described table, we can conclude that the major questions were asked from Algebra, Time & Work, Geometry, Data Interpretation, Profit/loss, Mensuration, and Trigonometry. The difficulty level of these questions was very high and questions were very time-consuming. Hence, we suggest you to give more time on these topics. Let us have a look of these questions-

    1) The internal bisector of the ∠B and ∠C of the ΔABC, intersect at O. If ∠A = 100, then the measure of ∠BOC is:-

    a) 110

    b) 130

    c) 140

    d) 120

    Ans. (c.)

    Explanation: in triangle ABC, 100 + 2x + 2y = 180; => x + y = 40;

    In triangle BOC,

    BOC = 180 – (x + y) = 180 – 40 = 140;

    2) A conical iron piece having diameter 28 cm and height 30cm is totally immersed in the rise of water level by 6.4cm. The diameter, in cm, of the vessel is:-

    a) 3.5

    b) 32

    c) 35

    d) 35/2

    Ans. (c.)

    Explanation: Let the radius of the vessel = r cms;

    Volume of displaced water = volume of conical iron piece;

    Pi* r2 * 6.4 = 1/3 * (14)2 * 30; => r = 17.5 cms;

    Hence, the diameter of the vessel = 35 cms.

    3) The value of the following is 3(Sin4Θ + Cos4Θ) + 2(Sin6Θ + Cos6Θ) + 12Sin2ΘCos2Θ

    a) 3

    b) 0

    c) 5

    d) 2

    Ans. (c.)

    Explanation:

    = 3(Sin4Θ + Cos4Θ) + 2(Sin6Θ + Cos6Θ) + 12Sin2ΘCos2Θ;

    = 3(Sin4Θ + Cos4Θ) + 2(Sin2Θ + Cos2Θ) (Sin4Θ + Cos4Θ - Sin2Θ Cos2Θ) + 12Sin2ΘCos2Θ;

    = 5(Sin4Θ + Cos4Θ) + 10Sin2ΘCos2Θ;

    = 5(Sin4Θ + Cos4Θ + 2Sin2ΘCos2Θ);

    = 5(Sin2Θ + Cos2Θ)2;

    =5;

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    4) If the area of the base, height and volume of a right prism be (3√3/ 2) P2 cm2, 100√3 cm and 7200 cm3 respectively, then the value of P will be?

    a) 2/√3

    b) √3

    c) 3/2

    d) 4

    Ans. (d.)

    Explanation: Volume of prism = area of base * height;

    7200 = (3√3/ 2) P2 * 100√3; => P = 4;

    5) If the discount of 10% is given on the marked price of a radio, the gain is 20%. If the discount is increased to 20%, the gain is:-

    a) 5%

    b) 6.67%

    c) 7.62%

    d) 6.25%

    Ans. (b.)

    Explanation: Suppose marked price = Rs. 100;

    Price after discount = Rs. 90;

    Since, the realized gain = 20%; Hence, the cost price = Rs. 75;

    If the discount price = 20%, then the price after discount = Rs. 80;

    Hence, the required answer = 5*100/75 = (20/3) %=6.67%;

    6) If 4a – 4/a + 3 = 0 then the value of a3 – 1/a3 + 3 =?

    a) 3/16

    b) 21/64

    c) 7/16

    d) 21/16

    Ans.

    Explanation: a – 1/a = -3/4;

    Cubing both sides-

    a3 – 1/a3 -3*( a – 1/a) = -27/64;

    a3 – 1/a3 = -27/64 + 3 *(-3/4);

    a3 – 1/a3 = -171/64;

    a3 – 1/a3 + 3 = -171/64 + 3 = 21/16;

    7) O is the circumcentre of ∆ABC. If ∠BAC = 85 ∠BCA = 75, the ∠OAC is equal to:-

    a) 60

    b) 70

    c) 50

    d) 40

    Ans. (b.)

    Explanation: Angle ABC = 180 – (85 +75) = 20;

    Hence, Angle OAC = 20 *2 = 40 degrees;

    Since, OA = OC; therefore, Angle OAC = Angle OCA;

    Hence, Angle OAC = 140/2 = 70 degrees;

    8) If A, B and C can complete a work in 6 days. If A can work twice faster than B thrice faster than C, then the number of days C alone can complete the work is:

    a) 22 Days

    b) 44 Days

    c) 33 Days

    d) 11 Days

    Ans. (c.)

    Explanation: (A + B + C)’s efficiency = (100/6)%;

    A’s efficiency = 2 * B’s efficiency = 3* C’s efficiency;

    3C + 3C/2 + C = 100/6; => 11C/2 = 100/6;

    C’s efficiency = 100/(3*11);

    Hence, C alone can finish the work in = 33 days;

    10) A circular swimming pool is surrounded by a concrete wall 4m wide. If the area of the concrete wall surrounding the pool is 11/25 that of the pool, then the radius (in m) of the pool is:

    a) 8

    b) 20

    c) 16

    d) 30

    Ans. (b.)

    Explanation: Let the radius of the swimming pool = R meter;

    Radius of pool with wall = (R + 4) meters;

    Pi* [(R + 4)2 – R2] = 11/25 * pi* R2;

    8(R + 2) = 11/25 * R2;

    After solving this equation- we get,

    R = 20 meter.

    11) Two pipes A and B can fill a tank with water in 30 minutes and 45 minutes respectively. The water pipe C can empty the tank in 36 minutes. First A and B are opened. After 12 minutes C is opened. Total time (in minutes) in which the tank will be filled up is:-

    a) 12

    b) 30

    c) 36

    d) 24

    Ans. (d.)

    Explanation: A’s efficiency to fill the tank in every minute= (100/30)% =3.33%

    B’s efficiency to fill the tank in every minute = (100/45)% = 2.22% and

    C’s efficiency to empty the tank in every minute = (100/36) = 2.78%;

    In 12 minutes, A and B will fill the tank in a minute = 5.55*12 = 66.6%;

    In 1 minute, A and B fill the tank = 5.55%;

    Therefore, A, B and C will fill the tank in 1 minute = 5.55 – 2.78 = 2.77%;

    Remaining tank to be filled = 100 -66.6 = 33.4%;

    Time taken to fill the empty tank = 33.4/2.77 = 12.05 minutes;

    Total time = 12 + 12 = 24 minutes;

    12) A shopkeeper allows a discount of 10% on the marked price of a camera. Marked price of the camera, which costs him Rs. 600, to make a profit of 20% should be:-

    a) Rs. 750

    b) Rs. 800

    c) Rs. 700

    d) Rs. 650

    Ans. (a.)

    Explanation: Cost price = Rs. 600;

    Suppose the marked price = Rs. X;

    X* 0.8 – 500 = 500* 20%;

    X= 500*1.2/0.8 = Rs. 750;

    13) O is the incentre of ∆PQR and ∠QPR = 50, then the measure of ∠QOR is :-

    a) 130

    b) 125

    c) 115

    d) 100

    Ans. (a.)

    Explanation: The inner radius will be perpendicular to the sides PQ and PR. Hence,

    In quadrilateral,

    APC + AOC = 180; => AOC = 130;

    Angle AOC = Angle QOR = 130 degrees;

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    14) If x2 + y2 + z2 = 2(x + z – 1), then the value of x3 + y3 + z3 =?

    a) 1

    b) 0

    c) 2

    d) -1

    Ans. (c.)

    Explanation: Take x =1 y =0 z =1
    1+0+1= 2
    2(1+1-1) =2
    So 1+0+1= 2;

    Directions: In Question nos. 16 to 19, Study the following bar-diagram and answer the questions.

    Electricity units consumed by a family in two consecutive years during July to November

    16) In how many months in 2012, the consumption of electric units was more than the average units consumption in that years.

    a) 5

    b) 2

    c) 3

    d) 4

    Ans. (b.)

    Explanation:  Average of electric units in the year 2012 = (600 + 700 + 400 + 300 + 200)/5 = 2200/5 = 440; hence, there will be only two months where consumption of units will be higher than the average i.e. july and august.

    17) The maximum difference in the units consumption between these two years has been found in the month of:-

    a) October

    b) August

    c) November

    d) July

    Ans. (c.)

    Explanation: In November, difference between the consumption of electric units = 500- 200 = 300; (which is higher than others)

    18) The average electric consumption by the family during these 5 months in 2013 is

    a) 470 units

    b) 460 units

    c) 450 units

    d) 400 units

    Ans. (b.)

    Explanation: Average of electric units in the year 2013 = (550 + 500 + 400 + 350 + 500)/5 = 2300/5 = 460.

    19) The Total units consumption in the year 2013 during these 5 months, in respect of the same in the previous year has been:-

    a) decreased by 2.27%

    b) increased by 4.54%

    c) increased by 2.27%

    d) found unaltered

    Ans. (b.)

    Explanation: The total electric units in the year 2013 = (550 + 500 + 400 + 350 + 500) = 2300;

    The total electric units in the year 2012 = (600 + 700 + 400 + 300 + 200) = 2200;

    The required percentage = 100*100/2200 = 4.54%;

    20) AC is a transverse common tangent to two circles with centers P and Q and radii 6cm and 3cm at the point A and C respectively. If AC cuts PQ at the point B and AB = 8cm then the length of PQ is:-

    a) 12 cm

    b) 13 cm

    c) 10 cm

    d) 15 cm

    Ans.

    Explanation: In triangle ABP, PB = 10 cms; (Pythagoras theorem)

    Triangle ABP and BQC are similar triangles; Hence,

    AP/QC = PB/BQ; => BQ = 5 cms;

    PQ = 10 + 5 = 15 cms;

    21) A dealer sold a bicycle at a profit of 10%. Had he brought the bicycle at 10% less price and sold it at a price Rs. 60 more, he would have gained 25%. The cost price of the bicycle was-

    a) Rs. 2600

    b) Rs. 2200

    c) Rs. 2400

    d) Rs. 2000

    Ans.

    Explanation: Cost price = Rs. x; SP = 1.1x;

    0.9*CP * 125%= 1.1x + 60;

    1.25*0.9*x = 1.1x + 60; => x = Rs. 2400;

    Directions: In Question nos. 22 to 24, The income of a state under different heads is given in the following pie-chart. Study the chart and answer the questions.

    22) The central angle of the sector representing income tax is

    a) 126°

    b) 119°

    c) 135°

    d) 150°

    Ans. (a.)

    Explanation: The required angle = 35% of 360 = 126 degrees.

    23) If the total income in a year be Rs. 733 crores then the income (in Rs. crores) from 'Income tax' and 'Excise duty' is:-

    a) Rs. 329.80

    b) Rs. 331.50

    c) Rs. 331.45

    d) Rs. 329.85

    Ans. (d.)

    Explanation: income from income tax and excise duty = (35 +10)% of 733 = 733*0.45= Rs. 329.85;

    24) If the income from the market tax in a year be Rs. 165 crores then the total income from other sources is (in Rs. crores):-

    a) Rs. 335

    b) Rs. 345

    c) Rs. 325

    d) Rs. 365

    Ans.

    Explanation: Market tax = 33% of the total tax;

    Hence, the total tax = 165 *100/33 = Rs. 500;

    The required total income = 67% of total tax = Rs. 335;

    25) A dealer buys an article listed at 100 and gets successive discounts of 10% and 20%. He spends 10% of the Cost Price on transportation. At what price should he sell the article to earn a profit of 15%?

    a) Rs. 91.20

    b) Rs. 92.00

    c) Rs. 90.80

    d) Rs. 91.08

    Ans.

    Explanation: Resultant successive discount = 20% + 10% - 20% * 10% = 28%;

    The buying price of the article= 100 * (100-28)% = 100*72% = Rs. 72;

    Buying price after transportation= 72 + 7.2 = Rs. 79.2;

    Hence, the selling price = 79.2 * 1.15 = Rs. 91.08;

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    26) A librarian purchased 50 story-books for his library. But he saw that he could get 14 more books by spending Rs. 76 more and the average price per book would be reduced by Rs. 1. The average price (in Rs.) of each book he bought, was:

    a) 25

    b) 15

    c) 20

    d) 10

    Ans.

    Explanation: Suppose, Total price of 50 story-books = Rs. x; Average price of 1 book = x/50;

    Total price of 64 books including 14 books = x + 76; Average price of 1 book = (x + 76)/64;

    (x+76)/64 = x/50 -1; => Rs. 500;

    Hence, The average price of 1 book = 500/50 = Rs. 10;

    28) The speed of a boat in still water is 6 Km/hr and the speed of stream is 1.5 Km/hr. A man rows to a place at a distance of 22.5 Km and comes back to the starting point. The total time taken by him is:-

    a) 6 hours

    b) 8 hours

    c) 10 hours

    d) 4 hours

    Ans. (b.)

    Explanation: Total time taken = 22.5/(6 + 1.5) + 22.5/(6 – 1.5) =  3 + 5 = 8 hours;

    29) A and B together can do a piece of work in 30 days. B and C together can do it in 20 days. A starts the work and works on it for 5 days, then B takes up and works for 15 days. Finally C finishes the work in 18 days. The number of days in which C alone can do the work when doing it seperately is:-

    a) 40 Days

    b) 24 Days

    c) 120 Days

    d) 60 Days

    Ans. (b.)

    Explanation: (A + B) ‘s efficieny = 100/30 = 3.33%;

    (B + C)’s efficiency = 100/20 = 5%;

    Assume that the efficiency of A, B and C = x%, y%, and z%;

    (x + y) = 3.33; (y + z) = 5;

    5x + 15y + 18z = 100;

    5(x + y) + 10(y + z) + 8z =100; => 5*3.33 + 10 * 5 + 8z =100;

    16.65 + 50 + 8z =100; => z= 4.168%;

    Hence, C can alone do that work = 100/4.168 = 24 days;

    30) If Tan A + Cot A = 2, then the value of Tan10A + Cot10A is-

    a) 1

    b) 210

    c) 2

    d) 4

    Ans.

    Explanation: Put A = 45; => 1 + 1 =2;

    Hence, Tan10A + Cot10A = (1)10 + (1)10 =2;

    31) The percentage of metals in a mine of lead ore is 60%. Now the percentage of silver is 3/4 % of metals and the rest is lead. If the mass of ore extracted from this mine is 8000kg, the mass (in kg.) of lead is:

    a) 4763

    b) 4762

    c) 4761

    d) 4764

    Ans. (d.)

    Explanation: The percentage of lead = 60 % of 8000 = 4800 kgs.

    The mass of silver = 4800 * ¾ % = 36 kgs.

    Hence, the mass of lead = 4800 – 36 = 4764 kgs;

    32) If a + b - c = 14 then the value of: 2b2c2 + 2c2a2 + 2a2b2 – a4 – b4 – c4

    a) 14

    b) 0

    c) 28

    d) 7

    Ans.

    Explanation: assume a= 7, b=7 and c=0;

    =2b2c2 + 2c2a2 + 2a2b2 – a4 – b4 – c4 = 0;

    33) ABCD is a cyclic quadrilateral. Diagnals AC and BD meets at P. If ∠APB = 110° and ∠CBD = 30°, then ∠ADB measures-

    a) 55°

    b) 80°

    c) 70°

    d) 30°

    Ans. (b.)

    Explanation: Angle CPB = 180 – 110 =70;

    Angle PCB = 180 – (30 + 70) = 80;

    Hence, angle ADB = angle PCB = 80; ( Angles subtended by a chord are equal);

    35) Average weight of 3 men A, B, C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E whose weight is 3kg more than that of D replaces A then the average weight of B, C, D and E becomes 79 kg. The weight of A in Kg is :

    a) 72

    b) 70

    c) 80

    d) 75

    Ans. (d.)

    Explanation: (A + B + C) = 84*3;

    (A + B + C + D)/4 = 80; => D = 320 – 252 = 68 kgs;

    The weight of E = 71 kgs;

    (B + C + D + E)/4 = 79; => B + C = 79*4 – 68 -71 = 177 kgs;

    Hence, the weight of A = 252 -177 = 75 kgs.

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    36) If x = z = 225 and y = 226 then the value of: x3 + y3 + z3 -3xyz =?

    a) 765

    b) 674

    c) 676

    d) 576

    Ans. (c.)

    Explanation: x3 + y3 + z3 -3xyz = ½ * (x + y + z)[(x-y)2 + (y –z)2 + (z –x)2];

    = ½ * (225 + 226 + 225) [(1)2 +(1)2-0] = 676*2/2= 676 ;

    37) AB and CD are two parallel chords of a circle lying on the opposite side of the center and the distance between them is 17cm. The length of AB and CD are 10 cm and 24 cm respectively. The radius(in cm) of the circle is:-

    a) 18

    b) 13

    c) 15

    d) 9

    Ans. (b.)

    Explanation: 122 + x2 = (17 –x)2 + 52; => x = 5 cms;

    Hence, the radius of the circle = √(12)2 +(5)2 = 13 cms;

    38) If 1 + Cos2Θ = 3SinΘCosΘ, then the integral value of CotΘ is (0<Θ< π/2)

    a) 0

    b) 2

    c) 1

    d) 3

    Ans. (c.)

    Explanation: Θ = 45 is satisfying the given equation-

    Hence, cot 45 =1;

    39) If two numbers are in the ratio 2 : 3 and the ratio becomes 3 : 4 when 8 is added to both the numbers, then the sum of the two numbers is:-

    a) 40

    b) 80

    c) 100

    d) 10

    Ans. (a.)

    Explanation: let numbers are 2x and 3x;

    (2x + 8)/(3x + 8) = &frac34; => 8x + 32 = 9x + 24;

    x = 8;  

    hence, the sum of these numbers = 5x = 40;

    40) Two towers A and B have lengths 45m and 15m respectively. The angle of elevation from the bottom of the B tower to the top of the A tower is 60°. If the angle of elevation from the bottom of A tower to the top of the B tower is Θ then value of Sin Θ is:-

    a) 1/2

    b) 1/√2

    c) √3/2

    d) 2/√3

    Ans. (a.)

    Explanation: tan60 = 45/CC’; => CC’ = 45/√3 ;

    tanΘ = 15/CC’ = 15/(45/√3) = 1/√3; => Θ = 30;

    Hence, sin Θ = &frac12;;

    41) In ∆ABC, D and E are two mid points of sides AB and AC respectively. If ∠BAC = 40° and ∠ABC = 65° then ∠CED is :-

    a) 125°

    b) 105°

    c) 75°

    d) 130°

    Ans. (b.)

    Explanation:  Triangle ADE and ABC are similar triangles.

    Hence, Angle ADE = Angle ABC = 65;

    Angle AED = Angle ACB = 180 – (40 + 65) = 75;

    Hence, Angle CED = 180 – 75 = 105;

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    42) If x + 1/x =1; then the value of: 2/(x2-x +2)=?

    a) 2/3

    b) 4

    c) 1

    d) 2

    Ans. (d.)

    Explanation: x + 1/x =1; => x2 –x +1 =0;

    x2 –x +2 =1;

    hence, the required answer = 2;

    43) The area of the triangle formed by the graphs of the equations x = 4, y = 3 and 3x + 4y = 12 is

    a) 12 sq. unit

    b) 6 sq. unit

    c) 4 sq. unit

    d) 3 sq. unit

    Ans. (b.)

    Explanation: The area of triangle ABC = &frac12; * 4 * 3 = 6 sq. units;

    45) Ram deposited a certain sum of money in a company at 12% per annum simple interest for 4 years and deposited equal amount in fixed deposit in a bank for 5 years at 15% per annum simple interest. If the difference in the interest from two sources is Rs. 1350 then the sum deposited in each case is :-

    a) Rs. 5000

    b) Rs. 4000

    c) Rs. 3000

    d) Rs. 6500

    Ans. (a.)

    Explanation: Suppose the principal amount be P.

    (P*5*15)/100  -  (P*12*4)/100  = 1350;

    75P - 48P = 135000;

    27P = 135000; => P = Rs. 5000;

    46) If SecΘ + TanΘ = 2 + √5, then the value of SinΘ is-

    a) 4/5

    b) 1/√5

    c) 2/√5

    d) √3/5

    Ans. (c.)

    Explanation: secΘ + tanΘ = √5 + 2;

    secΘ - tanΘ = √5 -2;

    secΘ = √5 ;

    sinΘ = 2/√5;

    47) A train leaves station A at 5 AM and reaches station B at 9 AM on the same day. Another train leaves station B at 7 AM and reaches station A at 10.30 AM on the same day. The time at which the two trains cross one another is:-

    a) 7.36 AM

    b) 8 AM

    c) 7.56 AM

    d) 8.26 AM

    Ans. (c.)

    Explanation: The time taken by first train to reach from A to B= 4 hours;

    The time taken by second train to reach from B to A= 3.5 hours;

    Let the distance between A and B = 100 kms.

    Hence, the speed of first train = 25 km/hr;

    The speed of second train = 100/3.5 = 28.57 km/hr;

    Distance travelled by first train in two hour = 50 kms.

    The time taken to meet each other = 50/53.57 = 0.93 hrs = 55.8 minutes = 56 minutes;

    Hence, the time to cross trains each other= 7:65 AM

    48) The value of the following is:

    Cos 24 + Cos 55 + Cos 125 + Cos 204 + Cos 300

    a) 1

    b) 2

    c) – 1/2

    d) &frac12;

    Ans. (d.)

    Explanation:

    =cos 24 + cos 55 + cos 125 + cos 204 + cos 300

    = cos 24 + cos 55 + cos (180-55) + cos (180+24) + cos (3600-60)

    = cos 24 + cos 55 - cos 24 - cos 55 + cos60

    = &frac12;

    8 myths about SSC exams preparation are finally busted

    DISCLAIMER: JPL and its affiliates shall have no liability for any views, thoughts and comments expressed on this article.

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